function genericIdentity<T>(arg: T): T {
return arg;
}
let myGenericIdentity: <U>(arg: U) => U = genericIdentity;
I see that the 'genericIdentity' function is accepting an argument of a generic type. However, I am unsure about how the assignment for 'myGenericIdentity' is functioning. Can you explain this further?
When you call the identity function, you need to specify a single unconstrained type parameter called Type. The function takes an argument arg of type Type and returns a value of the same type:
function identity<Type>(arg: Type): Type {
return arg;
}
//IntelliSense suggests:
//function identity<Type>(arg: Type): Type
You can find out the type of identity by using TypeScript's typeof type operator:
type TypeofIdentity = typeof identity
// type TypeofIdentity = <Type>(arg: Type) => Type
The type of identity is expressed as
<Type>(arg: Type) => TypeNote that Type and arg are just placeholders for the actual types and arguments used in the function, and you can rename them without changing the function's type. So, these declarations are equivalent to the original:
type TypeofIdentity = <T>(x: T) => T
Or this:
type TypeofIdentity = <Input>(arg: Input) => Input
This means that your myIdentity variable has been assigned the exact same type as identity, indicating they are equivalent functions. You can create variables with the same type, like so:
var x: typeof identity;
var x: typeof myIdentity; // allowed
var x: <T>(x: T) => T; // also valid
In essence, creating myIdentity is similar to creating a new variable with the same value and assigning it to another variable of the same type:
const date = new Date(); // make a thing
let myDate: Date = date; // assign thing to another variable
Therefore, you can use myIdentity in the same way as identity since they have identical functionalities:
const str = myIdentity("str");
// const str: "str"
const tru = myIdentity(true);
// const tru: true
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